Find the probability that at most 5 defective bolts will be found in a box of 200 bolts if it is known that 2 per cent of such bolts are expected to be defective .(you may take the distribution to be Poisson; e-4= 0.0183).
0.031
To find the probability that at most 5 defective bolts will be found in a box of 200 bolts, given that 2% of such bolts are expected to be defective, we can use the Poisson distribution.
The Poisson distribution formula is given by:
P(X = k) = (e^(-λ) * λ^k) / k!
Where:
P(X = k) is the probability of getting exactly k defective bolts,
e is the mathematical constant approximately equal to 2.71828,
λ is the average number of defective bolts expected (λ = np),
k is the number of defective bolts we want to find.
In this case, we have n = 200 bolts and the expected proportion of defective bolts p = 2%.
So, λ = np = 200 * 0.02 = 4.
Now, we can calculate the probability of finding at most 5 defective bolts (P(X ≤ 5)) using the Poisson distribution:
P(X ≤ 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)
Using the given value for e^(-4) = 0.0183, we can calculate the probability step-by-step:
P(X = 0) = (e^(-4) * 4^0) / 0! = 0.0183 * 1 / 1 = 0.0183
P(X = 1) = (e^(-4) * 4^1) / 1! = 0.0183 * 4 / 1 = 0.0732
P(X = 2) = (e^(-4) * 4^2) / 2! = 0.0183 * 16 / 2 = 0.1464
P(X = 3) = (e^(-4) * 4^3) / 3! = 0.0183 * 64 / 6 = 0.0976
P(X = 4) = (e^(-4) * 4^4) / 4! = 0.0183 * 256 / 24 = 0.0976
P(X = 5) = (e^(-4) * 4^5) / 5! = 0.0183 * 1024 / 120 = 0.0732
Adding these probabilities together, we get:
P(X ≤ 5) = 0.0183 + 0.0732 + 0.1464 + 0.0976 + 0.0976 + 0.0732 = 0.5063
Therefore, the probability that at most 5 defective bolts will be found in a box of 200 bolts is approximately 0.5063, or 50.63%.
To find the probability that at most 5 defective bolts will be found in a box of 200 bolts, given that 2% of such bolts are expected to be defective, we can use the Poisson distribution.
The Poisson distribution is commonly used to model the number of events that occur in a fixed interval of time or space. It is characterized by a single parameter, λ (lambda), which represents the average number of events in the interval.
In this case, λ would be the expected number of defective bolts in a box of 200 bolts, which can be calculated by multiplying the total number of bolts (200) by the probability of a bolt being defective (2% or 0.02):
λ = 200 * 0.02 = 4
Now, we can use the Poisson distribution formula to calculate the probability:
P(X ≤ 5) = e^(-λ) * (λ^0/0!) + e^(-λ) * (λ^1/1!) + e^(-λ) * (λ^2/2!) + e^(-λ) * (λ^3/3!) + e^(-λ) * (λ^4/4!) + e^(-λ) * (λ^5/5!)
Given that e^-4 = 0.0183, we can substitute this value into the formula:
P(X ≤ 5) = 0.0183 * (4^0/0!) + 0.0183 * (4^1/1!) + 0.0183 * (4^2/2!) + 0.0183 * (4^3/3!) + 0.0183 * (4^4/4!) + 0.0183 * (4^5/5!)
Now, we can simplify and calculate the probability:
P(X ≤ 5) = 0.0183 * 1 + 0.0183 * 4 + 0.0183 * 8 + 0.0183 * 64/6 + 0.0183 * 256/24 + 0.0183 * 1024/(24 * 5)
P(X ≤ 5) = 0.0018 + 0.0732 + 0.1464 + 0.0586 + 0.0762 + 0.0765
P(X ≤ 5) ≈ 0.4337
So, the probability of finding at most 5 defective bolts in a box of 200 bolts is approximately 0.4337.
Here's one way to do this problem.
Poisson distribution (m = mean):
P(x) = e^(-m) m^x / x!
Note: mean = .02*200 = 4
Find P(0) through P(5). Add together for the probability.
Here's P(0):
P(0) = (e^ -4) (4^0) / (0!)
= (0.0183) (1) / (1)
= 0.0183
I hope this will help get you started.